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What does the inverse of the embedding process mean exactly? For example the embedding process is:

  1. Sequentially extract out every 8-bit data from watermark-bit-stream.
  2. Obtain a random number, generated by pseudo random system, which points to one of n blocks of host image.
  3. Embed extracted the 8-bit watermarking data into the 8 lower-band coefficients in the block pointed by previous step.
  4. Repeat step 1 to step 3, until the watermark bit stream is run out.
  5. The proposed employee replace bit to embedded watermark bit stream, and it was hidden at position bit 3 in the selected 8-bit coefficient. If the watermark bit is “1” then bit 3 to “1” otherwise “0”.

So the inverse of this process would be starting from no. 5-1???Is this right?? any help would be appreciated thanks

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closed as not a real question by Randolph West, Daniel Andersson, Renan, Journeyman Geek, Daniel Beck Jul 22 '12 at 13:53

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1 Answer 1

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Step 5 is not a step, it is just a description of the process of embedding.

So the process is to add each byte of watermark to a random block in the original image. So in order to detect the watermark later, you need to know the seed that was used and use the same random number generator.

It doesn't say whether it detects collisions, so there is potential for a block to have a part of the water mark applied twice, so you'd need to run in reverse. In order words, generate a list of random numbers to get to the last block that would have been processed (equal to the length of the watermark) do the test or de-embedding, then step back through each random number to the beginning, to the first block processed.

It also doesn't say exactly what process to perform on the block but presumably it is reversable.

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So to clarify the inverse steps is 1. generate a list of random numbers ... 2. do the test or de-embedding 3. step back through each random number to the beginning to the first block prcessed. Have I got the inverse process right?? –  stbb24 May 15 '12 at 6:16
    
@stbb24 You have to understand that the description you have provided is too vague to allow for accurate answers. However, in principle, the answer provided should be right. Note that the "random" number list in the inverse process is identical to the original random number list, so not really random anymore :) –  Paul May 15 '12 at 6:30
    
I know it's not that accurate and sorry for that I just want to know if I understood the inverse part correctly :) –  stbb24 May 15 '12 at 6:47

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